The answer is #0.62L#.

Dilution calculations allow us to determine how a solute's concentration is reduced in solution usually by adding more solvent to the mix. This can be expressed mathematically by

#C_1V_1 = C_2V_2#, where

#C_1# = initial concentration (or molarity );

#V_1# = initial volume;

#C_2# - final concentration (or molarity);

#V_2# - final volume;

We can determine the final concentration of #NaCl# solution by

#C_2 = V_1/V_2 * C_1 = (150 * 10^(-3) L)/(2.5 L) * 6.5M = 0.39 M#

We know that molarity is defined as

#C = n_(solute)/(V_(solution)#, which gives us #n_(solute) = C * V_(solution) = 0.39 * 2.5 = 0.975 mol es# in #2.5 L#.

However, we were given #m_(NaCl) = 13.8 g#, which translates, knowing that #NaCl#'s molar mass is equal to #58.5 g/(mol)#, into

#n = (13.8 g)/(58.5 g/(mol)) = 0.24 mol es# This is roughly #1/4#th of the number of moles present in #2.5L#, so we would expect the final volume to be roughly #1/4#th of #2.5L -> 0.63L#.

#V = n/C = (0.24 mol es)/(0.39 M) = 0.62L#, which is close to our estimate.